Solution 20.3

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(New page: We start by converting the speed to SI units. <math>60\text{ km/h }=\frac{60\times 1000}{60\times 60}=\frac{50}{3}=\text{16}\textrm{.}\text{7 m}{{\text{s}}^{\text{-1}}}</math> The diagra...)
Current revision (13:20, 12 October 2010) (edit) (undo)
 
Line 14: Line 14:
& F\le \mu R \\
& F\le \mu R \\
& 5556\le \mu \times 9800 \\
& 5556\le \mu \times 9800 \\
 +
& \\
& \mu \ge \frac{5556}{9800} \\
& \mu \ge \frac{5556}{9800} \\
-
& \mu \ge 0.567 \\
+
& \mu \ge 0\textrm{.}567 \\
\end{align}</math>
\end{align}</math>

Current revision

We start by converting the speed to SI units.

\displaystyle 60\text{ km/h }=\frac{60\times 1000}{60\times 60}=\frac{50}{3}=\text{16}\textrm{.}\text{7 m}{{\text{s}}^{\text{-1}}}

The diagram shows the forces on the car.

Image:20.3.gif

\displaystyle F=m\frac{{{v}^{2}}}{r}=1000\times \frac{{{\left( \frac{50}{3} \right)}^{2}}}{50}=5556\text{ N}

\displaystyle R=mg=9800

\displaystyle \begin{align} & F\le \mu R \\ & 5556\le \mu \times 9800 \\ & \\ & \mu \ge \frac{5556}{9800} \\ & \mu \ge 0\textrm{.}567 \\ \end{align}